Let $ABC$ be an isosceles triangle with $AB = AC$. Let $AD$ be the diameter of the circumcircle of $ABC$ and let $P$ be a point on the smaller arc $BD$. The line $DP$ intersects the rays $AB$ and $AC$ at points $M$ and $N$, respectively. The line $AD$ intersects the lines $BP$ and $CP$ at points $Q$ and $R$, respectively. Prove that the midpoint of $MN$ lies on the circumcircle of $PQR$
Problem
Source: The 1st SAFEST Olympiad 2019 p1
Tags: geometry, midpoint, circumcircle, isosceles
MathMonster0923
11.09.2019 12:20
Since $AB = AC$ and $AD$ is diameter, $AD$ is the perpendicular bisector of $BC$.
$C$ is just a reflection of $B$ across $AD$.
Claim 1: $CD$ is the angle bisector of $\angle RCQ$.
Proof: $\angle QCD = \angle QBD = \angle PBD = \angle PCD = \angle RCD$. $\Box$
Claim 2: $QDCN$ and $MBRD$ are cyclic.
Proof: $\angle QDN = \angle ADP = \angle ACP = \angle ACR = \angle ACD - \angle RCD = \angle DCN - \angle QCD = \angle QCN$.
$\angle RDM = \angle QDN = \angle QCN = \angle ACR = \angle ABR$. $\Box$
From Claim 2, $\angle MRD = \angle MBD = 90^{\circ}$ and $\angle DQN = 180^{\circ} - \angle DCN = 90^{\circ}$
Let $(PQR)$ intersect $MN$ at a second time at $X$. We will show $X$ is the midpoint of $MN$.
Claim 3: $PX$ is the angle bisector of $\angle QPR$.
Proof: $\angle QPX = \angle BPM = 90^{\circ} - \angle BPA = 90^{\circ} - \angle BDA = \angle DBC = \angle DPC = \angle XPR$. $\Box$
From Claim 3, $\angle XRQ = \angle XPQ = \angle RPX = \angle RQX$. This means $XR = XQ$ and $X$ is on the perpendicular bisector of $RQ$.
Construct point $Y$ such that $MRQY$ is a rectangle.
Let the perpendicular bisector of $RQ$ intersect $MY$ at $Z$. Since $MRQY$ is a rectangle, $Z$ is the midpoint of $MY$.
$ZX$ and $YN$ are both perpendicular to $RQ$, so $ZX || YN$.
So the homothety centered at $M$ bringing $Z$ to $Y$ also brings $X$ to $N$.
Therefore, $\frac{MX}{XN} = \frac{MZ}{ZY} = 1$. $X$ is the midpoint of $MN$. $\blacksquare$
Jjesus
13.06.2020 19:14
parmenides51 wrote: What is it SAFEST? Is from some country?
parmenides51
13.06.2020 19:15
South AFrica + ESTonia TST
Jjesus
13.06.2020 20:28
parmenides51 wrote: South AFrica + ESTonia TST Thank you!
Ricochet
13.06.2020 22:07
https://artofproblemsolving.com/community/c6h2078586p14944688
Kimchiks926
15.06.2020 15:18
Since $AR$ is perpendicular bisector of $BC$, we get that $\angle BAR = \angle RAC = \alpha $. Claim 1: $NAPQ$ is cyclic Proof: Note that: $$ \angle BAR = \angle BAD = \angle MPB = \angle QPN = \angle RAC = \angle QAN = \alpha $$ Claim 2: $MAPR$ is cyclic Proof: Note that $\angle BPC = \angle BPR = 180 - 2\alpha $. Therefore: $$ \angle MAR + \angle MPR =\angle MAR+ \angle MPB + \angle BPR = \alpha + \alpha + 180 -2\alpha =180$$ From Claim 1 it follows that $\angle APD =\angle AQN = 90 $, which implies that $QN \parallel BC$. Let $QN \cap AB = X$. Since $AQ$ is perpendicular bisector of $BC$ and $QN \parallel BC$, we get that $Q$ is midpoint of $XN$. Now let $Y$ denote midpoint of $MN$, then $QY$ is midline in $\triangle XMN$, therefore $QY \parallel XM \parallel AM$. To finish the problem note that: $$ \angle AMP = \angle PRQ = \angle DYQ = \angle PYQ $$which implies that $PRYQ$ is cyclic and we are done. $\blacksquare$